# Closed loop equation initial value

I am doing math for a closed loop equation. I know the transfer function for a closed loop system is:

$$V_{out} = G \times V_{in} \tag {Eq 1}$$

$$G = \frac {A}{1 + \beta A_0}$$

$$\ A \$$ - open loop gain
$$\ \beta \$$ - Feedback factor.

Now when I do the math like

$$V_{out} = A(V_{in} - \beta V_{out}) \tag {Eq 2}$$

What is the initial of Vout to start with?

Vout numbers with 1 & 2 do not match and initial value of Vout is making a lot of difference.

Why are they not matching?

If A = 1000, B = 0.25, Vin =1, Vout =3.98;

>From Eq 1

Vout = 1000/(1+0.25*1000) = 3.98

From Eq2

Vout_initial = 0

Vout = 1000*(1-0.25 x 0) = 1000

Vout = 1000*(1-0.25 x 1000) = -249000

Vout = 1000*(1-0.25 x -249000) = 62251000

Never settles, what am I doing wrong?

If I chose A = 1; B =0.1, Eq1 and Eq2 numbers match eventually to 0.91

Vout = 1*(1-0.1 x 0) = 1

Vout = 1*(1-0.1 x 1) = 0.9

Vout = 1*(1-0.1 x 0.9) = 0.91

and

Vout = 1/(1+0.1*1) = 0.909

What's going on? What fundamentals am I missing?

• A schematic might be helpful. There's a button on the editor toolbar. Double-click a component to edit its properties. Hit Save and Insert when done. You don't need a CircuitLab account when run from SE. – Transistor Mar 19 at 23:29
• Try read this electronics.stackexchange.com/questions/441184/… – G36 Mar 19 at 23:49
• It's not an iterative equation. – Chu Mar 19 at 23:51
• @Aakash, I fixed some missing ( ) in Eq. 1 and converted a couple of lines to MathJAX which I suspect you'll love. You can continue on with the remainder of the post, if you wish. – Transistor Mar 19 at 23:56
• Thanks @Transistor, great job, this looks so nice :) – Aakash Mar 19 at 23:57

## 1 Answer

You're trying to form an iterative equation. There are many ways of going about it, sometimes it works, sometimes it doesn't.

This way works:

$$\small V_o=\frac{A}{1+AB}V_i$$

add $$\\small V_o\$$ to both sides:

$$\small 2V_o=\frac{A}{1+AB}V_i +V_o$$

Divide by 2: $$\small V_o=\frac{1}{2}\left(\frac{A}{1+AB}V_i +V_o\right)$$ Make it iterative: $$\small V_o[k]=\frac{1}{2}\left(\frac{A}{1+AB}V_i +V_o[k-1]\right)$$

Let $$\\small A=1000; B=0.25; V_i=1\$$ $$\small V_o[k]=\frac{1}{2}\left(\frac{1000}{251} +V_o[k-1]\right)$$ hence:

$$\small V_o[k]=\frac{500}{251} +\frac{V_o[k-1]}{2}$$

Let $$\\small V_o[0]=0\$$, then solve iteratively:

$$\small V_o[1]=1.992$$ $$\small V_o[2]=2.988$$ $$\small V_o[3]=3.486$$ $$\small V_o[4]=3.735$$ This converges to $$\\small V_o[k]=3.984\$$ at 17 iterations.

Interestingly(?), a square root algorithm can be formulated in a similar way:

Let $$\\small y=\sqrt{x}\$$ $$\small y^2=x$$ $$\small y=\frac{x}{y}$$ $$\small 2y=\frac{x}{y}+y$$ $$\small y=\frac{1}{2}\left(\frac{x}{y}+y\right)$$ Make iterative: $$\small y[k]=\frac{1}{2}\left(\frac{x}{y[k-1]}+y[k-1]\right)$$

e.g. let $$\\small x=9,\:y[0]=1\$$ $$\small y[1]=\frac{1}{2}\left(\frac{9}{1}+1\right)=5$$ $$\small y[2]=\frac{1}{2}\left(\frac{9}{5}+5\right)=3.4$$ $$\small y[3]=\frac{1}{2}\left(\frac{9}{3.4}+3.4\right)=3.0235$$ $$\small y[4]=\frac{1}{2}\left(\frac{9}{3.0235}+3.0235\right)=3.00009$$

• "There are many ways of going about it, sometimes it works, sometimes it doesn't", what you are looking for is a contraction map (like the new formulation for the iteration you found), and the use of such contraction map is to find a fixed point of the system $f(x^*)=x^*$. If $f()$ is a contraction mapping "any" initial guess converges to the fixed-point. – jDAQ Mar 20 at 1:05
• @Chu..wow... thank you so much. . – Aakash Mar 20 at 2:01
• What @Chu is pointing out is not how the system will behave, but a way to calculate $V_{out}$. As you have said yourself, the output will be $V_{out}(t) = V_{in}(t) \frac {A}{1 + \beta A_0}$. Your second "method", comes from your perception that $V_{out} = A(V_{in} - \beta V_{out})$ is different than $V_{out}(t) = V_{in}(t) \frac {A}{1 + \beta A_0}$ (and it is not), you do not have to guess $V_{out}(t)$ at any moment, just calculate it using $V_{in}$. – jDAQ Mar 20 at 2:51
• Cool, thanks guys, I think the math works just fine. I am curious what will happen to control loop Vo at the instant an Vi in applied. There must be some delay before the loop can react. What is this delay called? How do I learn about this delay? – Aakash Mar 20 at 3:21
• There needn't be a delay there (in the model), specially for Op Amp and similarly fast circuits, modelling that delay and the behavior at the very moment you start applying $V_i$ might show that it quickly goes to the value the equation tells you, probably without even oscillating. But if you are interested in bigger delays that actually create instabilities and odd behaviors, look at discrete-time systems and users.ece.utexas.edu/~buckman/H3.pdf – jDAQ Mar 20 at 5:13